Subjective Type

If $$A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 5 & 3 \\ 0 & 2 & 1 \end{bmatrix} $$ find $$A^{-1}$$, using elementary row transformations.

Solution

For using elementary row transformations, we take
$$A= IA$$, where $$I$$ is identity matrix of order $$3 \times 3$$
$$\begin{bmatrix}1 & -1 & 0 \\2 & 5 & 3 \\0 & 2 & 1\end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix} A$$
Applying $$R_2 \rightarrow R_2 - 2R_1$$, we get
$$\begin{bmatrix}1 & -1 & 0 \\0 & 7 & 3 \\0 & 2 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\-2 & 1 & 0 \\0 & 0 & 1\end{bmatrix} A$$
Applying, $$R_2 \rightarrow R_2 - 3R_3$$, we get
$$\begin{bmatrix}1 & -1 & 0 \\0 & 1 & 0 \\0 & 2 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\-2 & 1 & -3 \\0 & 0 & 1\end{bmatrix} A$$
Applying, $$R_1 \rightarrow R_1 + R_2$$ and $$R_3 \rightarrow R_3 - 2R_2$$,
$$\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix} = \begin{bmatrix}-1 & 1 & -3 \\-2 & 1 & -3 \\4 & -2 & 7\end{bmatrix} A$$

And we know that $$A^{-1}A=I$$
Hence
$$A^{-1} = \begin{bmatrix}-1 & 1 & -3 \\-2 & 1 & -3 \\4 & -2 & 7\end{bmatrix}$$

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